What are the solutions to the equation 6k^2+2k+9=-3 ?

The solutions to the equation 6k^2+2k+9=-3 are k=-1/6+i(sqrt(71))/6 ,k=-1/6-i(sqrt(71))/6

Find the zeros of 6k^2+2k+9=-3

The zeros of 6k^2+2k+9=-3 are k=-1/6+i(sqrt(71))/6 ,k=-1/6-i(sqrt(71))/6

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6 k^{2} + 2 k + 9 = - 3

k = - \frac{1}{6} + i \frac{71}{6}, k = - \frac{1}{6} - i \frac{71}{6}

Solution steps

6 k^{2} + 2 k + 9 = - 3

Move

3

to the left side

6 k^{2} + 2 k + 12 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

Simplify

2^{2} - 4 \cdot 6 \cdot 12 : 271 i

k_{1, 2} = \frac{- 2 \pm 2 71 i}{2 \cdot 6}

Separate the solutions

k_{1} = \frac{- 2 + 2 71 i}{2 \cdot 6}, k_{2} = \frac{- 2 - 2 71 i}{2 \cdot 6}

k = \frac{- 2 + 2 71 i}{2 \cdot 6} : - \frac{1}{6} + i \frac{71}{6}

k = \frac{- 2 - 2 71 i}{2 \cdot 6} : - \frac{1}{6} - i \frac{71}{6}

The solutions to the quadratic equation are:

k = - \frac{1}{6} + i \frac{71}{6}, k = - \frac{1}{6} - i \frac{71}{6}

x^{2} - 4 x = 52

- 4 u < - 16

\frac{( x - 1 )}{( x + 2 ) ( x + 5 )} \leq 0

(5 - 3 x) - (- 4 x + 6) = (8 x + 1) - 3 (2 x + 3)

7 x < 21 or x - 4 > 1

What are the solutions to the equation 6k^2+2k+9=-3 ?
The solutions to the equation 6k^2+2k+9=-3 are k=-1/6+i(sqrt(71))/6 ,k=-1/6-i(sqrt(71))/6
Find the zeros of 6k^2+2k+9=-3
The zeros of 6k^2+2k+9=-3 are k=-1/6+i(sqrt(71))/6 ,k=-1/6-i(sqrt(71))/6